Electric Potential: Practice Problem & Solution
Three concentric spherical shells have radii $a$, $b$ and $c$ ($a < b < c$) and have surface charge densities $\sigma$, $-\sigma$ and $\sigma$ respectively. If $V_A$, $V_B$ and $V_C$ denote the potentials of the three shells, then for $c = a + b$, we have: (2009)
Solution Explained:
To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:
Potential of shell A is $V_A = \frac{\sigma}{\epsilon_0}(a - b + c)$. Potential of shell C is $V_C = \frac{\sigma}{\epsilon_0}\left(\frac{a^2 - b^2}{c} + c\right)$. Given $c = a + b$, $V_C = \frac{\sigma}{\epsilon_0}\left(\frac{(a-b)(a+b)}{a+b} + c\right) = \frac{\sigma}{\epsilon_0}(a - b + c) = V_A$. Thus $V_A = V_C \neq V_B$.
Leave a Reply